Free Cheatsheet · Topic E.4 · SL + HL

IB Physics HL E4 Fission — Complete Cheatsheet

Every formula, definition and exam trap for IB Physics Topic E.4 — fission, chain reactions, the multiplication factor, critical mass, reactor components and energy calculations. Hand-built by an IBO-experienced Singapore tutor.

Topic: Fission (Quantum & Nuclear) Syllabus: Topic E.4 (SL + HL) Read time: ~12 minutes Last updated: Apr 2026

Topic E.4 — Fission — is the IB Physics nuclear topic with the strongest real-world hook. Half of the marks come from describing reactor components (moderator, control rods, fuel, coolant, containment) and the other half from quantitative calculations of fission energy via mass defect and $E = \Delta m c^2$. The HL extension layers in critical mass and multiplication-factor reasoning, and Paper 2 questions almost always close with an evaluation of nuclear power's environmental trade-offs.

This cheatsheet condenses every formula, reaction equation, trick and trap from Topic E.4 SL + HL onto one page. Scroll to the bottom for the printable PDF, the full notes pack and the gated tutorial library used by Photon Academy students in Singapore.

Cross-reference with the official IB Physics subject page for the full syllabus.

§1 — Nuclear Fission E.4 SL & HL

Definition

Fission: a heavy nucleus absorbs a slow (thermal) neutron and splits into two lighter nuclei (fission fragments) plus 2–3 neutrons and a large amount of energy.

$${}^{1}_{0}n + \,{}^{235}_{92}\text{U} \rightarrow \,{}^{141}_{56}\text{Ba} + \,{}^{92}_{36}\text{Kr} + 3\,{}^{1}_{0}n$$

Conservation laws

In every nuclear reaction:

  • Nucleon number $A$ is conserved.
  • Proton number $Z$ is conserved.
  • Mass-energy is conserved (the missing mass appears as kinetic energy of the products).
TrickAlways check both $A$ and $Z$ when balancing fission equations. If two unknowns appear, set up simultaneous equations from the $A$ and $Z$ balance.
TrapDo NOT confuse "fission" (heavy $\rightarrow$ light) with "fusion" (light $\rightarrow$ heavy). Both release energy because both move toward higher binding energy per nucleon — but in opposite directions on the BE/A curve.

Binding energy & fission

Fission releases energy because the products have higher BE/nucleon than U-235.

Energy released:$E = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}}$

U-235: BE/A $\approx 7.6$ MeV. Fragments: BE/A $\approx 8.4$ MeV. $\Rightarrow E \approx 200$ MeV per fission.

NoteIron-56 has the highest BE/nucleon ($\approx 8.79$ MeV). Nuclei on either side can release energy by moving toward Fe-56 — heavier nuclei via fission, lighter nuclei via fusion.
n ²³⁵U ¹⁴¹Ba ⁹²Kr 2–3 n n + ²³⁵U → ¹⁴¹Ba + ⁹²Kr + 3n + ~200 MeV

Nuclear fission. A slow neutron splits a U-235 nucleus into two lighter fragments, releasing 2–3 neutrons and ~200 MeV — the fragments are more tightly bound than the original.

From the Photon question bank

What examiners actually test on this concept:

n + U-235 → fragments + 2–3nConserve A and Z~200 MeV per fissionProducts higher BE/A
Fission: a heavy nucleus absorbs a thermal neutron and splits into two fragments + 2–3 neutrons + ~200 MeV. Balance $A$ (top) and $Z$ (bottom). frequent
Energy is released because the fragments have HIGHER binding energy per nucleon than U-235 (moving toward Fe-56).
~200 MeV = $3.2\times10^{-11}$ J per fission — use it when exact masses aren't given.

§2 — Chain Reactions E.4 SL & HL

Multiplication factor $k$

Definition:$k = \dfrac{\text{neutrons in generation } (n+1)}{\text{neutrons in generation } n}$
  • $k < 1$: subcritical — reaction dies out.
  • $k = 1$: critical — steady-state reactor.
  • $k > 1$: supercritical — exponential growth (bomb behaviour).
TrickFor $k$: count neutrons available for fission only. Subtract those lost (absorbed by control rods or non-fissile material) and those escaping through the walls from the total produced.

Critical mass (HL)

Critical mass = minimum mass for a sustained chain reaction ($k \geq 1$). It is determined by a competition between two effects:

  • Fission rate $\propto$ volume $\propto r^3$.
  • Neutron leakage $\propto$ surface area $\propto r^2$.

Larger mass $\Rightarrow$ smaller surface-to-volume ratio $\Rightarrow$ smaller leakage fraction $\Rightarrow$ chain reaction can be sustained.

TrapCritical mass is NOT the mass at which fission starts. Fission can occur in a single atom. Critical mass is about sustaining the chain reaction.
124 k > 1 ⇒ each generation grows (supercritical)

Chain reaction. Each fission releases neutrons that trigger further fissions; when $k>1$ the number grows generation by generation.

From the Photon question bank

What examiners actually test on this concept:

Multiplication factor kSubcritical / critical / supercriticalCritical massSurface vs volume
$k$ = neutrons in the next generation ÷ this one: $k<1$ subcritical (dies out), $k=1$ critical (steady reactor), $k>1$ supercritical (bomb). frequent
Critical mass = the minimum mass to SUSTAIN the chain — not to start fission (a single atom can fission).
Bigger mass → smaller surface/volume → less neutron leakage → sustainable.

§3 — Nuclear Reactor Components E.4 SL & HL

ComponentFunctionExample materials
Fuel rodsContain the fissile material; site of fissionEnriched U-235 ($\sim$3–5%)
ModeratorSlows fast neutrons to thermal energies via elastic collisionsWater (H$_2$O), heavy water (D$_2$O), graphite
Control rodsAbsorb neutrons to regulate the reaction; fully inserted = shutdownBoron, cadmium, hafnium
Heat exchangerTransfers thermal energy from coolant to a secondary water circuit (steam)Pressurised water, CO$_2$
ContainmentShields the environment from radiationConcrete, lead, steel
TrickA good moderator must be (i) light (similar mass to a neutron $\Rightarrow$ maximum energy transfer per elastic collision) and (ii) a poor neutron absorber (so the neutrons are slowed but not lost).
TrapThe statement "the heat exchanger slows neutrons" is WRONG. The moderator slows neutrons. The heat exchanger only transfers thermal energy from the primary coolant to the secondary water circuit to produce steam.
core (moderator) fuelcontrol hot coolant heat exch. cool steam turbine generator

Reactor schematic. Fuel and control rods sit in a moderator; hot coolant carries thermal energy to a heat exchanger that raises steam to drive a turbine and generator.

From the Photon question bank

What examiners actually test on this concept:

Moderator slows neutronsControl rods absorbHeat exchanger ≠ moderatorGood moderator = light + poor absorber
The moderator SLOWS fast neutrons to thermal energies (elastic collisions); control rods ABSORB neutrons to regulate $k$. frequent
The heat exchanger only transfers thermal energy to the secondary water circuit — it does NOT slow neutrons (a common wrong statement).
A good moderator is light (maximum energy transfer per collision) and a poor neutron absorber.

§4 — Energy Calculations E.4 SL & HL (quantitative HL)

Mass defect:$\Delta m = Z m_p + (A - Z) m_n - m_{\text{nucleus}}$
Energy from mass:$E = \Delta m \cdot c^2$
Useful conversion:$1\,\text{u} \equiv 931.5$ MeV
Power from mass loss:$P = \dot{m}\, c^2$
Power from rate:$P = R \cdot E_{\text{fission}}$
TrickWhen using atomic masses (not nuclear masses), the electron masses cancel in the subtraction for $\Delta m$, so atomic masses can be used directly — provided you are consistent on both sides.

Energy density comparison

FuelEnergy / MJ kg${}^{-1}$
Coal$\approx 30$
Oil$\approx 45$
Enriched U (3%)$\approx 5 \times 10^6$
Pure U-235$\approx 8 \times 10^7$

Nuclear fuels are roughly $10^5$–$10^6$ times more energy-dense than fossil fuels.

Note$E \approx 200$ MeV per U-235 fission $= 3.2 \times 10^{-11}$ J. Use this if the exact masses are not given.
Coal 30 Oil 45 U (3%) U-235 nuclear ~10⁵–10⁶× fossil fuels MJ kg⁻¹ (log)

Energy density. Uranium fuel packs roughly $10^5$–$10^6$ times more energy per kilogram than coal or oil.

From the Photon question bank

What examiners actually test on this concept:

E = Δm·c²1 u = 931.5 MeVP = R·E_fissionEnergy density
$E=\Delta m\,c^2$ with $1\,\text{u}=931.5$ MeV; scale to power via $P=R\,E_{\text{fission}}$ ($R$ = fissions per second) or $P=\dot m c^2$. frequent (HL)
Nuclear fuel is ~$10^5$–$10^6$× more energy-dense than fossil fuels.
Using atomic masses, the electron masses cancel in $\Delta m$ if you're consistent on both sides.

§5 — Safety & Environment E.4 SL & HL

Advantages

  • Very high energy density.
  • Low CO$_2$ during operation.
  • Reliable baseload power (weather-independent).
  • Small land footprint.

Disadvantages / risks

  • Long-lived radioactive waste (HLW — high-level waste).
  • Meltdown / accident risk.
  • Thermal pollution (warm water released to aquatic ecosystems).
  • High construction cost and long build times.
  • Nuclear weapons proliferation risk.
  • Uranium mining damage.
Trap"Nuclear power produces no pollution" is WRONG in IB exams. It produces radioactive waste and thermal pollution. Say: low carbon emissions during operation, NOT pollution-free.
From the Photon question bank

What examiners actually test on this concept:

Low carbon in operationRadioactive wasteNot 'pollution-free'Reliable baseload
Say "LOW carbon emissions during operation", never "pollution-free" — it produces long-lived radioactive waste and thermal pollution. frequent
Advantages: very high energy density, reliable weather-independent baseload, small land footprint.
Risks: high-level waste, meltdown, thermal pollution, cost/build time, and proliferation.

Five-step method for fission energy calculations (HL)

  1. Write the balanced equation and identify all species.
  2. Find $\Delta m = m_{\text{reactants}} - m_{\text{products}}$ (in u).
  3. Convert: $E = \Delta m \times 931.5$ MeV (or $\times c^2$ in SI).
  4. Convert to joules if needed: multiply MeV by $1.602 \times 10^{-13}$.
  5. Scale to power: $P = R \times E_{\text{fission}}$ or $P = \dot{m}\, c^2$.

Worked Example — Reactor Power Output

Question (HL Paper 2 style — 7 marks)

A pressurised-water reactor (PWR) outputs $P = 1.0$ GW of thermal power from the fission of U-235. Assume each fission releases $E_{\text{fission}} = 200$ MeV.
(a) Calculate the number of fissions per second $R$ in the reactor. [3]
(b) Hence find the rate of mass loss $\dot{m}$ of U-235 (i.e. mass converted to energy per second). [2]
(c) State and explain whether the multiplication factor $k$ is held above, below or exactly at 1 during steady operation. [2]

Solution

  1. Convert: $E_{\text{fission}} = 200 \times 10^6 \times 1.602 \times 10^{-19} = 3.20 \times 10^{-11}$ J.  (M1)
  2. $R = P / E_{\text{fission}} = (1.0 \times 10^9)/(3.20 \times 10^{-11})$.  (M1)
  3. $R \approx 3.13 \times 10^{19}$ fissions per second.  (A1)
  4. Mass converted: $\dot{m} = P / c^2 = (1.0 \times 10^9)/(3.00 \times 10^8)^2$.  (M1)
  5. $\dot{m} \approx 1.11 \times 10^{-8}$ kg s${}^{-1}$ $\approx 11$ μg per second of mass converted to energy.  (A1)
  6. $k = 1$ exactly.  (A1)
  7. Each fission must produce, on average, exactly one further fission — so neutron production matches absorption + leakage. Higher $k$ would be supercritical (power runaway); lower $k$ would shut the reactor down.  (R1)

Examiner's note: Two common errors. (i) Confusing the rate of mass conversion ($\dot{m} = P/c^2$, on the order of micrograms per second) with the rate of fuel consumption (kilograms per day) — only a tiny fraction of the fuel mass actually becomes energy. (ii) Saying $k$ "stays close to 1" — for steady operation it must be exactly 1; control rods continuously trim it to that value.

Common Student Questions

Why does fission release energy?
Because the fission fragments (medium-mass nuclei, $A \approx 90$–$145$) sit higher on the binding-energy-per-nucleon curve than U-235. U-235 has BE/A $\approx 7.6$ MeV; the fragments have BE/A $\approx 8.4$ MeV. The difference of $\sim 0.8$ MeV per nucleon, multiplied by 235 nucleons, gives roughly 200 MeV released per fission.
What does the multiplication factor $k$ tell you?
$k$ is the ratio of neutrons in one generation to neutrons in the previous generation. $k < 1$: subcritical (chain dies out); $k = 1$: critical (steady reactor power); $k > 1$: supercritical (power grows exponentially — bomb behaviour). Reactors are designed to sit just above $k = 1$ and are trimmed back to exactly 1 by the control rods.
What is the difference between the moderator and the control rods?
The moderator slows fast neutrons to thermal energies via elastic collisions, so they can cause fission in U-235. Materials must be light (water, heavy water, graphite) and poor neutron absorbers. Control rods absorb neutrons to regulate the reaction rate; fully inserted = shutdown. Materials are good neutron absorbers like boron, cadmium or hafnium. Saying that the control rods "slow neutrons" is a guaranteed mark loss.
Does critical mass mean the mass at which fission starts?
No — fission can occur in a single atom. Critical mass is the minimum mass needed to sustain a chain reaction ($k \geq 1$). Below the critical mass, too many neutrons leak out through the surface (which scales as $r^2$) compared to the number produced inside the volume (scaling as $r^3$). Larger mass means a smaller surface-to-volume ratio and therefore less leakage.
Is nuclear power "pollution-free"?
No — that is a guaranteed wrong answer in IB exams. Nuclear power has very low carbon emissions during operation, but it produces radioactive waste, thermal pollution (warm water released to rivers/sea), uranium-mining damage, and carries proliferation and meltdown risks. The correct phrasing is: low carbon emissions during operation, NOT pollution-free.

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